Question: An arithmetic sequence with first term $1$ has a common difference of $6$. A second sequence begins with $4$ and has a common difference of $7$. In the range of $1$ to $100$, what is the largest number common to both sequences?
Answer: Let $a$ be the smallest common term. We know that \begin{align*}
a & \equiv 1 \pmod 6\\
a & \equiv 4 \pmod 7
\end{align*} We see that $a \equiv 1 \pmod 6$ means that there exists a non-negative integer $n$ such that $a=1+6n$. Substituting this into $a \equiv 4 \pmod 7$ yields \[1+6n\equiv 4\pmod 7\implies n\equiv 4\pmod 7\] So $n$ has a lower bound of $4$. Then $n\ge 4\implies a=1+6n\ge 25$. We see that $25$ satisfies both congruences so $a=25$. If $b$ is any common term, subtracting $25$ from both sides of both congruences gives \begin{align*}
b-25 & \equiv -24\equiv 0\pmod 6 \\
b-25 & \equiv -21\equiv 0\pmod 7
\end{align*} Since $\gcd(6,7)=1$, we have $b-25\equiv 0\pmod {6\cdot 7}$, that is, $b\equiv 25\pmod{42}.$ So $b=25+42m$ for some integer $m$. The largest such number less than $100$ is $\boxed{67}$, which happens to satisfies the original congruences.